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JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    Consider line \[L\equiv \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}\]. Point P (1, 0, 0) and Q are such that PQ is perpendicular to line L and the mid-point of PQ lies on the line L then Q is

    A)  (3,-4, -2)                             

    B)  (5,-8,-4)

    C)  (1,-1,-10)                           

    D)  (2, -3, 8)

    Correct Answer: B

    Solution :

    Q is reflection of P w.r.t. to given line L. Let Q is \[(\alpha ,\beta ,\gamma )\]then \[(\alpha ,\beta ,\gamma )\]lies on then line \[\Rightarrow \]\[\frac{\frac{\alpha +1}{2}-1}{2}=\frac{\frac{\beta }{2}+1}{-3}=\frac{\frac{\gamma }{2}+10}{8}=\lambda \]and\[2(\alpha -1)-3\beta +8\gamma =0\]\[\Rightarrow \]\[\alpha =5,\beta =-8,\gamma =-4.\]


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